Question

Find the equation of the circle with center at (2, -3) and radius 5.

• A. \( x^2 + y^2 - 4x + 6y - 12 = 0 \)
• B. \( x^2 + y^2 - 4x + 6y + 12 = 0 \)
• C. \( x^2 + y^2 + 4x - 6y - 12 = 0 \)
• D. \( x^2 + y^2 + 4x - 6y + 12 = 0 \)

Hint:

For the equation of a circle, use \( (x - h)^2 + (y - k)^2 = r^2 \), then expand to the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

Answer 1

The Correct Option is A

The equation of a circle with center \( (h, k) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Given center \( (2, -3) \) and radius \( r = 5 \): \[ (x - 2)^2 + (y - (-3))^2 = 5^2 \] \[ (x - 2)^2 + (y + 3)^2 = 25 \] Expand: \[ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 25 \] \[ x^2 + y^2 - 4x + 6y + 13 - 25 = 0 \] \[ x^2 + y^2 - 4x + 6y - 12 = 0 \] Thus, the equation of the circle is: \[ \boxed{x^2 + y^2 - 4x + 6y - 12 = 0} \]